(2+x)(3x-8)=0

Solution for (2+x)(3x-8)=0 equation:

(2+x)(3x-8)=0

We add all the numbers together, and all the variables

(x+2)(3x-8)=0

We multiply parentheses ..

(+3x^2-8x+6x-16)=0

We get rid of parentheses

3x^2-8x+6x-16=0

We add all the numbers together, and all the variables

3x^2-2x-16=0

a = 3; b = -2; c = -16;
Δ = b2-4ac
Δ = -22-4·3·(-16)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-14}{2*3}=\frac{-12}{6}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+14}{2*3}=\frac{16}{6}
=2+2/3
$